# Margin of Error

How to find the margin of error from the sample proportion: definition, formula, 1 example, and its solution.

## Sample Proportion

### Definition

In reality,

we don't know the p (proportion)

of the whole data (population).

So, to find the p,

we pick a sample

and find the proportion of the sample:

p^.

p^ is read as [p hat].

## Formula

### Formula

But p^ and p cannot be always the same.

The sample cannot always represent

the whole population.

Then people found out that,

under a certain confidence level,

p is in this interval.

p^ - Z√[p^⋅q^]/n ≤ p ≤ p^ + Z√[p^⋅q^]/n

Shortly, p is in

p^ ± Z√[p^⋅q^]/n.

This Z√[p^⋅q^]/n

is the margin of error.

Z: Z-score (related to the confidence level)

p^: Probability (proportion) of the wanted event

q^ = 1 - p^

n: Number of the values of the sample

As you can see,

if n (sample number) increases,

the margin of error decreases.

So we can determine p (population proportion)

more precisely.

But if n increases,

you need more resources (time, money)

to find p.

So it's good to find the right n.

## Example

### Example

### Solution

It says

in a random sample,

64% of the people favored the candidate A.

So the sample proportion is

p^ = 0.64.

p^ = 0.64

Then q^ = 1 - 0.64.

This q^ means

the sample proportion of the people

not favored the candidate A.

Probability: Not A

1 - 0.64 = 0.36.

The sample number is 256.

So n = 256.

p^ = 0.64

q^ = 0.36

n = 256

Then [p^⋅q^]/n = [0.64⋅0.36]/256.

Multiply 100^{2}

to both of the numerator and the denominator.

Then [64⋅36]/[256⋅100^{2}].

64 = 2^{6}

36 = 2^{2}⋅3^{2}

256 = 2^{8}

Prime Factorization

Cancel 2^{6}⋅2^{2} in the numerator

and cancel the denominator 2^{8}.

3^{2}/100^{2} = [3/100]^{2}

Power of a Quotient

3/100 = 0.03

Find the z-score.

It says

P(0 ≤ Z ≤ 1.96) = 0.4750.

Recall that

for the normal distribution curve,

the left side and the right side

are the same.

So P(-1.96 ≤ Z ≤ 1.96) = 0.4750⋅2 = 0.95.

P(-1.96 ≤ Z ≤ 1.96) = 0.95 means

the z-score for 95% is Z = 1.96.

So Z = 1.96.

Then the margin of error, ME,

is equal to,

the z-score, 1.96

times

square root, [p^⋅q^]/n, 0.03^{2}.

So the margin of error is

(ME) = 1.96⋅√0.03^{2}.

√0.03^{2} = 0.03

Square Root

1.96⋅0.03 = 0.0588

0.0588 = 5.88%

So the margin of error is 5.88%.

This means

for 95% probability,

the candidate A's favorability rating is

64% ± 5.88%:

58.12% ~ 69.88%.